Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AIME I Problems/Problem 8"
(→‎Solution 1)
Line 27: Line 27:
  
 
We can extend <math>AB</math> and <math>AC</math> to <math>̈B'</math> and <math>C'</math> such that circle <math>\omega_A</math> is the incircle of <math>\triangle AB'C'</math>.
 
We can extend <math>AB</math> and <math>AC</math> to <math>̈B'</math> and <math>C'</math> such that circle <math>\omega_A</math> is the incircle of <math>\triangle AB'C'</math>.
 +
 +
== Diagram ==
 +
<asy>
 +
unitsize(0.3cm);
 +
draw(circle((0,0),18));
 +
pair A = (9 * sqrt(3), -9);
 +
pair B = (-9 * sqrt(3), -9);
 +
pair C = (0,18);
 +
draw(A--B--C--cycle);
 +
draw(circle((0,-6),12), gray);
 +
draw(circle((3*sqrt(3),3),12), gray);
 +
draw(circle((-3*sqrt(3),3),12), gray);
 +
 +
pair X = (0, 3-sqrt(117));
 +
pair Y = ( (sqrt(351)-sqrt(27))/2, (sqrt(117)-3)/2 );
 +
pair Z = ( (sqrt(27) - sqrt(351))/2, (sqrt(117)-3)/2 );
 +
dot(X);
 +
dot(Y);
 +
dot(Z);
 +
 +
draw(X--Y--Z--cycle, dashed);
 +
</asy>
  
 
==Solution 2==
 
==Solution 2==

Revision as of 15:16, 18 February 2022

Problem

Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega$. Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A$, $\omega_B$, and $\omega_C$ meet in six points$-$two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC$, and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC$. The side length of the smaller equilateral triangle can be written as $\sqrt{a}-\sqrt{b}$, where $a$ and $b$ are positive integers. Find $a+b$.

Diagram

[asy] unitsize(0.3cm); draw(circle((0,0),18)); pair A = (9 * sqrt(3), -9); pair B = (-9 * sqrt(3), -9); pair C = (0,18); draw(A--B--C--cycle); draw(circle((0,-6),12), gray); draw(circle((3*sqrt(3),3),12), gray); draw(circle((-3*sqrt(3),3),12), gray); pair X = (0, 3-sqrt(117)); pair Y = ( (sqrt(351)-sqrt(27))/2, (sqrt(117)-3)/2 ); pair Z = ( (sqrt(27) - sqrt(351))/2, (sqrt(117)-3)/2 ); dot(X); dot(Y); dot(Z); draw(X--Y--Z--cycle, dashed); [/asy]

Solution 1

We can extend $AB$ and $AC$ to $̈B'$ (Error compiling LaTeX. Unknown error_msg) and $C'$ such that circle $\omega_A$ is the incircle of $\triangle AB'C'$.

Diagram

[asy] unitsize(0.3cm); draw(circle((0,0),18)); pair A = (9 * sqrt(3), -9); pair B = (-9 * sqrt(3), -9); pair C = (0,18); draw(A--B--C--cycle); draw(circle((0,-6),12), gray); draw(circle((3*sqrt(3),3),12), gray); draw(circle((-3*sqrt(3),3),12), gray); pair X = (0, 3-sqrt(117)); pair Y = ( (sqrt(351)-sqrt(27))/2, (sqrt(117)-3)/2 ); pair Z = ( (sqrt(27) - sqrt(351))/2, (sqrt(117)-3)/2 ); dot(X); dot(Y); dot(Z); draw(X--Y--Z--cycle, dashed); [/asy]

Solution 2

Let bottom left point as the origin, the radius of each circle is $36/3=12$, note that three centers for circles are $(9\sqrt{3},3),(12\sqrt{3},12),(6\sqrt{3},12)$

It is not hard to find that one intersection point lies on $\frac{\sqrt{3}x}{3}$ since the intersection must lie on the angle bisector of the bigger triangle, plug it into equation $(x-9\sqrt{3})^2+(\frac{\sqrt{3}x}{3}-3)^2=12^2$, getting that $x=\frac{15\sqrt{3}+3\sqrt{39}}{2}$, the length is $2*(\frac{15\sqrt{3}+3\sqrt{39}-18\sqrt{3}}{2}=3\sqrt{39}-3\sqrt{3}$, leads to the answer $378$

~bluesoul

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_8&oldid=171411"